Rotational Dynamics of a light cup

[asura]

Homework Statement

A common physics demonstration consists of a ball resting at the end of a board of length L that is elevated at an angle of theta with the horizontal. A light cup is attached to the board at Rc (Rc is a distance up the board from the bottom, it is not past the support stick) so that it will catch the ball when the support stick is suddenly removed. a) show that the ball will lag behind the falling board when theta < 35.3 and b) the ball will fall into the cup when the board is supported at this limiting angle and the cup is placed at
Rc = (2L)/(3cos$$\theta$$)

Homework Equations

$$\alpha$$r = a
x = v_ot + .5at²

The Attempt at a Solution

The ball falls straight down from the end of the board, so I found the time that it takes with x = v_ot + .5at² .Next I wanted to find the time it takes for the beam (and cup) to fall down, but I'm not sure how to do this. I tried using a component of gravity (gcos$$\theta$$) as the tangential acceleration, but then I realized that this value changes as the beam falls. So, how do I approach this problem?

 

[Doc Al]

Start by finding the angular acceleration of the stick as a function of angle.

 

[thomate1]

I would approach this problem by first identifying the key variables and forces at play. In this case, we have the angle of elevation (theta), the length of the board (L), the distance of the cup from the bottom of the board (Rc), and the acceleration due to gravity (g). The key forces acting on the system are the downward force of gravity and the normal force of the board on the ball.

To solve part a), we can use the equation \alphar = a, where \alpha is the angular acceleration and r is the distance from the axis of rotation (in this case, the point where the board is supported). We can also use the equation x = vot + .5at2, where x is the distance the ball falls, vo is the initial velocity (which is zero in this case), a is the tangential acceleration, and t is the time.

To find the time it takes for the board (and cup) to fall, we can use the equation \alpha = \frac{a}{r}, where a is the tangential acceleration and r is the distance from the axis of rotation. We can then use the equation \theta = \frac{1}{2} \alpha t^2 to find the time it takes for the board to fall to an angle of theta. We can then use this time in the equation x = vot + .5at2 to find the distance the board (and cup) falls.

To solve part b), we can use the equation \alphar = a to find the tangential acceleration at the limiting angle of theta = 35.3 degrees. We can then use this value in the equation x = vot + .5at2 to find the distance the ball falls before reaching the cup. We can then set this distance equal to Rc and solve for Rc to find the distance at which the cup should be placed for the ball to fall into it.

In conclusion, by using the appropriate equations and considering the key variables and forces at play, we can solve for the time and distance at which the ball will lag behind the falling board and the distance at which the cup should be placed for the ball to fall into it. This demonstrates the principles of rotational dynamics and can be a fun and engaging physics demonstration.